Sampling Distributions

1.1 What proportion of this bowl’s balls are red?

Take a look at the bowl in the following Figure. It has a certain number of red and a certain number of white balls all of equal size. Furthermore, it appears the bowl has been mixed beforehand as there does not seem to be any particular pattern to the spatial distribution of red and white balls.

Let’s now ask ourselves, what proportion of this bowl’s balls are red?

One way to answer this question would be to perform an exhaustive count: remove each ball individually, count the number of red balls and the number of white balls, and divide the number of red balls by the total number of balls. However this would be a long and tedious process.

Observe that ____ of the balls are red and there are a total of ____ balls and thus ___ % of the shovel’s balls are red. We can view the proportion of balls that are red in this shovel as a guess of the proportion of balls that are red in the entire bowl. While not as exact as doing an exhaustive count, our guess of ___% took much less time and energy to obtain.

However, say, we started this activity over from the beginning. In other words, we replace the 50 balls back into the bowl and start over. Would we remove exactly 17 red balls again? In other words, would our guess at the proportion of the bowl’s balls that are red be exactly 34% again? Maybe?

What if we repeated this exercise several times? Would I obtain exactly 17 red balls each time? In other words, would our guess at the proportion of the bowl’s balls that are red be exactly 34% every time? Surely not.

Let’s try do this on the computer…

To this end, we use a data frame bowl in the moderndive package whose rows correspond exactly with the contents of the actual bowl.

head(bowl)

# A tibble: 6 x 2
  ball_ID color
    <int> <chr>
1       1 white
2       2 white
3       3 white
4       4 red  
5       5 white
6       6 white

# View(bowl) # Use this in the console

Observe in the output that bowl has ___ rows, telling us that the bowl contains ___ equally-sized balls. The first variable ball_ID is used merely as an “identification variable”, none of the balls in the actual bowl are marked with numbers. The second variable color indicates whether a particular virtual ball is red or white.

Now that we have a virtual analogue of our bowl, we now need a virtual analogue for the shovel seen in Figure 2; we’ll use this virtual shovel to generate our virtual random samples of 50 balls. We’re going to use the rep_sample_n() function included in the moderndive package. This function allows us to take repeated, or replicated, samples of size n. Run the following and explore.

virtual_shovel <- bowl %>% 
  rep_sample_n(size = 50)

virtual_shovel

# A tibble: 50 x 3
# Groups:   replicate [1]
   replicate ball_ID color
       <int>   <int> <chr>
 1         1    1804 red  
 2         1     308 red  
 3         1    1663 white
 4         1     345 white
 5         1    2125 white
 6         1    2025 red  
 7         1    2204 white
 8         1     813 white
 9         1     352 red  
10         1    1962 white
# … with 40 more rows

Next we can find out how many res ones are there in our virtual_shovel

virtual_shovel %>% 
  summarize(num_red = sum(color=="red"))

# A tibble: 1 x 2
  replicate num_red
      <int>   <int>
1         1      20

How about the proportion on red? We can use the mutate (new) function to create a new variable, in this case prop_red.

virtual_shovel %>% 
  summarize(num_red = sum(color == "red")) %>% 
  mutate(prop_red = num_red / 50)

# A tibble: 1 x 3
  replicate num_red prop_red
      <int>   <int>    <dbl>
1         1      20      0.4

1.2 Using the virtual shovel many times

virtual_samples <- bowl %>% 
  rep_sample_n(size = 50, reps = 30)

kable(virtual_samples)

replicate	ball_ID	color
1	1614	red
1	2146	white
1	841	white
1	130	red
1	1456	red
1	1100	white
1	742	white
1	1423	white
1	203	red
1	1631	red
1	271	white
1	860	white
1	1321	white
1	2314	white
1	284	white
1	497	white
1	373	white
1	219	white
1	200	red
1	736	red
1	1759	white
1	1993	white
1	2139	white
1	929	red
1	303	red
1	2140	white
1	460	white
1	52	red
1	2074	red
1	1912	white
1	2344	red
1	470	red
1	866	white
1	2025	red
1	1589	red
1	365	white
1	646	white
1	627	red
1	1616	red
1	521	white
1	258	red
1	601	red
1	1900	white
1	2113	white
1	2061	white
1	2104	white
1	640	white
1	1024	red
1	2370	white
1	1876	white
2	1970	white
2	2218	white
2	1782	white
2	795	white
2	771	white
2	182	white
2	1408	white
2	386	white
2	2225	white
2	179	red
2	837	red
2	1244	red
2	842	red
2	1299	red
2	22	white
2	1864	white
2	2198	white
2	1084	white
2	1827	red
2	2071	red
2	1303	red
2	427	red
2	990	white
2	653	white
2	1349	white
2	1137	white
2	1527	white
2	1471	red
2	157	red
2	1167	red
2	1461	red
2	2117	white
2	1369	red
2	66	red
2	2268	red
2	1590	white
2	2335	white
2	2021	white
2	1402	white
2	2341	white
2	1373	red
2	2072	white
2	2115	white
2	924	red
2	2248	red
2	2127	white
2	1340	white
2	1319	white
2	1374	red
2	1106	white
3	203	red
3	1690	white
3	232	red
3	1485	white
3	1892	white
3	555	white
3	396	red
3	817	white
3	1966	red
3	2315	white
3	1064	white
3	830	red
3	785	white
3	2314	white
3	462	red
3	826	red
3	1144	red
3	606	red
3	1782	white
3	1468	white
3	1867	white
3	972	red
3	717	white
3	1524	white
3	2053	red
3	312	white
3	2274	red
3	641	white
3	853	white
3	841	white
3	98	red
3	1942	white
3	1003	white
3	805	white
3	1797	white
3	1905	white
3	854	red
3	374	white
3	2225	white
3	627	red
3	1149	white
3	1330	white
3	1400	white
3	833	red
3	947	white
3	73	white
3	1105	white
3	1339	white
3	1178	white
3	1992	red
4	563	white
4	1499	red
4	692	white
4	203	red
4	1035	white
4	588	red
4	1453	red
4	329	white
4	2043	red
4	298	red
4	110	white
4	284	white
4	1031	red
4	1929	red
4	2044	white
4	302	white
4	344	white
4	81	white
4	1080	red
4	1148	red
4	564	white
4	1321	white
4	2017	white
4	2258	white
4	840	white
4	143	red
4	1974	red
4	2265	red
4	1602	red
4	1176	red
4	371	white
4	663	red
4	565	red
4	1980	white
4	342	white
4	784	white
4	414	red
4	1616	red
4	1553	white
4	245	white
4	2020	red
4	2254	white
4	490	white
4	1753	red
4	1178	white
4	1829	white
4	1559	white
4	1713	white
4	553	white
4	2334	red
5	1655	red
5	426	white
5	228	white
5	2067	red
5	605	white
5	1791	white
5	529	white
5	1494	red
5	905	white
5	1295	white
5	360	red
5	1802	white
5	1090	white
5	2261	red
5	1403	white
5	869	white
5	2189	white
5	1205	red
5	2203	white
5	1151	white
5	902	red
5	87	red
5	633	red
5	859	white
5	1077	white
5	819	red
5	1705	white
5	2068	red
5	1271	red
5	473	white
5	1220	white
5	2188	red
5	2093	white
5	92	white
5	2029	white
5	1658	red
5	1477	red
5	734	white
5	1379	white
5	736	red
5	1374	red
5	2367	white
5	2341	white
5	2202	white
5	828	white
5	1913	white
5	1952	white
5	1660	white
5	995	white
5	2088	white
6	2209	red
6	474	red
6	64	red
6	1324	red
6	1728	red
6	1606	white
6	1831	white
6	1387	white
6	252	white
6	1856	white
6	1768	white
6	620	red
6	1228	red
6	24	white
6	1885	red
6	1021	red
6	697	white
6	1240	white
6	2125	white
6	1053	red
6	1209	white
6	2070	white
6	2186	red
6	2052	white
6	650	white
6	827	red
6	1313	red
6	731	white
6	2371	white
6	1257	white
6	1452	red
6	1451	red
6	869	white
6	363	white
6	2193	red
6	1017	red
6	668	red
6	568	red
6	240	red
6	1572	white
6	479	white
6	1195	red
6	1485	white
6	267	white
6	329	white
6	1527	white
6	622	red
6	291	white
6	414	red
6	2372	white
7	1297	white
7	1704	red
7	2091	red
7	1602	red
7	964	white
7	209	red
7	158	red
7	2031	white
7	587	white
7	2198	white
7	1932	white
7	1940	red
7	1415	red
7	2203	white
7	2130	white
7	1527	white
7	791	white
7	1939	white
7	2175	white
7	75	red
7	1670	white
7	1529	red
7	1813	white
7	328	white
7	1787	white
7	1810	red
7	416	white
7	1900	white
7	1692	white
7	530	red
7	367	red
7	129	red
7	741	white
7	1245	white
7	177	red
7	2225	white
7	920	red
7	1906	red
7	68	white
7	483	white
7	2147	white
7	240	red
7	1509	white
7	2053	red
7	2088	white
7	1554	white
7	270	red
7	128	white
7	86	white
7	382	white
8	2348	red
8	937	red
8	142	white
8	2039	white
8	1239	white
8	1291	white
8	833	red
8	1015	white
8	608	white
8	1458	white
8	2292	white
8	155	red
8	1107	white
8	1264	white
8	1153	white
8	1542	white
8	746	white
8	447	red
8	1248	white
8	189	white
8	871	red
8	464	white
8	1442	white
8	423	white
8	57	white
8	152	red
8	276	red
8	787	red
8	11	white
8	17	red
8	2168	white
8	675	white
8	1978	white
8	396	red
8	2340	red
8	1706	white
8	1359	red
8	1067	white
8	2029	white
8	676	red
8	1111	white
8	1889	red
8	1510	white
8	24	white
8	228	white
8	2161	red
8	2202	white
8	1412	white
8	1672	white
8	1225	white
9	139	white
9	883	red
9	1732	white
9	2198	white
9	2248	red
9	2391	white
9	1182	white
9	1501	white
9	630	white
9	652	white
9	1814	white
9	63	white
9	38	white
9	2397	red
9	1750	white
9	1319	white
9	1939	white
9	2269	red
9	1394	red
9	1016	red
9	543	red
9	1076	white
9	90	white
9	1738	red
9	1305	white
9	423	white
9	557	white
9	2311	white
9	1927	red
9	183	red
9	446	white
9	1057	white
9	779	red
9	103	white
9	1960	white
9	43	white
9	1124	white
9	1021	red
9	1168	red
9	1724	white
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9	595	white
9	979	white
9	1854	white
9	268	white
9	2027	red
9	855	red
9	1502	white
9	404	white
9	783	white
10	2140	white
10	2095	white
10	2220	red
10	776	white
10	1615	white
10	1606	white
10	244	white
10	1910	white
10	1182	white
10	1975	white
10	1481	white
10	1973	white
10	372	white
10	336	white
10	279	white
10	1741	red
10	1639	white
10	658	red
10	1163	white
10	938	red
10	846	red
10	2201	red
10	841	white
10	571	white
10	891	red
10	1836	white
10	640	white
10	804	white
10	1596	red
10	1324	red
10	1907	white
10	987	white
10	2361	white
10	727	white
10	1500	white
10	1835	red
10	762	white
10	1934	white
10	1321	white
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12	70	white
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Observe that while the first 50 rows of replicate are equal to 1, the next 50 rows of replicate are equal to 2. This is telling us that the first 50 rows correspond to the first sample of 50 balls while the next 50 correspond to the second sample of 50 balls. This pattern continues for all reps = 30 replicates and thus virtual_samples has \(30 \times 50 = 1500\) rows.

virtual_prop_red <- virtual_samples %>% 
  group_by(replicate) %>% 
  summarize(red = sum(color == "red")) %>% 
  mutate(prop_red = red / 50)

virtual_prop_red

# A tibble: 30 x 3
   replicate   red prop_red
       <int> <int>    <dbl>
 1         1    20     0.4 
 2         2    20     0.4 
 3         3    17     0.34
 4         4    22     0.44
 5         5    17     0.34
 6         6    23     0.46
 7         7    19     0.38
 8         8    16     0.32
 9         9    15     0.3 
10        10    18     0.36
# … with 20 more rows

#kable(virtual_prop_red) # To see all 30 samples

Let’s visualize the distribution of these 33 proportions red based on 33 virtual samples using a histogram with binwidth = 0.05

ggplot(virtual_prop_red, aes(x = prop_red)) +
  geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white", fill = "steelblue") +
  labs(x = "Proportion of 50 balls that were red", 
       title = "Distribution of 30 proportions red")

Observe that occasionally we obtained proportions red that are less than ____, while on the other hand we occasionally we obtained proportions that are greater than ____. However, the most frequently occurring proportions red out of 50 balls were between ____ % and ____ % (for ___ out 30 samples). Why do we have these differences in proportions red? Because of ___________________.

Exercise 1.1 Redo the above activity with 1000 repeated samples and state your conclusions.

1.2.1 Using different shovels

If your goal was still to estimate the proportion of the bowl’s balls that were red, which shovel would you choose? Why? Let’s try to answer these questions.

# Segment 1: sample size = 25 ------------------------------
# 1.a) Virtually use shovel 1000 times
virtual_samples_25 <- bowl %>% 
  rep_sample_n(size = 25, reps = 1000)

# 1.b) Compute resulting 1000 replicates of proportion red
virtual_prop_red_25 <- virtual_samples_25 %>% 
  group_by(replicate) %>% 
  summarize(red = sum(color == "red")) %>% 
  mutate(prop_red = red / 25)

# 1.c) Plot distribution via a histogram
p1 <- ggplot(virtual_prop_red_25, aes(x = prop_red)) +
  geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white") +
  labs(x = "Pro of 25 balls that were red", title = "25") 

# Segment 2: sample size = 50 ------------------------------
# 2.a) Virtually use shovel 1000 times
virtual_samples_50 <- bowl %>% 
  rep_sample_n(size = 50, reps = 1000)

# 2.b) Compute resulting 1000 replicates of proportion red
virtual_prop_red_50 <- virtual_samples_50 %>% 
  group_by(replicate) %>% 
  summarize(red = sum(color == "red")) %>% 
  mutate(prop_red = red / 50)

# 2.c) Plot distribution via a histogram
p2 <- ggplot(virtual_prop_red_50, aes(x = prop_red)) +
  geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white") +
  labs(x = "Pro of 50 balls that were red", title = "50")  

# Segment 3: sample size = 100 ------------------------------
# 3.a) Virtually using shovel with 100 slots 1000 times
virtual_samples_100 <- bowl %>% 
  rep_sample_n(size = 100, reps = 1000)

# 3.b) Compute resulting 1000 replicates of proportion red
virtual_prop_red_100 <- virtual_samples_100 %>% 
  group_by(replicate) %>% 
  summarize(red = sum(color == "red")) %>% 
  mutate(prop_red = red / 100)

# 3.c) Plot distribution via a histogram
p3 <- ggplot(virtual_prop_red_100, aes(x = prop_red)) +
  geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white") +
  labs(x = "Pro of 100 balls that were red", title = "100") 


plot_grid(p1, p2, p3, nrow = 1)

Observe that as the sample size increases, the ______ of the 1000 replicates of the proportion red decreases. In other words, as the sample size increases, there are less differences due to sampling variation and the distribution centers more tightly around the same value. Eyeballing the above Figure, things appear to center tightly around roughly ____%.

# n = 25
virtual_prop_red_25 %>% 
  summarize(sd = sd(prop_red))

# A tibble: 1 x 1
      sd
   <dbl>
1 0.0944

# n = 50
virtual_prop_red_50 %>% 
  summarize(sd = sd(prop_red))

# A tibble: 1 x 1
      sd
   <dbl>
1 0.0681

# n = 100
virtual_prop_red_100 %>% 
  summarize(sd = sd(prop_red))

# A tibble: 1 x 1
      sd
   <dbl>
1 0.0478

Number of slots in shovel	Standard deviation of proportions red
25	0.0978
50	0.0669
100	0.0471

As the sample size increases our numerical measure of spread decreases; there is less variation in our proportions red. In other words, as the sample size increases, our guesses at the true proportion of the bowl’s balls that are red get more consistent and precise.

1.3 What did we learn?

This was our first attempt at understanding two key concepts relating to sampling for estimation:

The effect of sampling variation on our estimates.
The effect of sample size on sampling variation.

Let’s now introduce some terminology and notation as well as statistical definitions related to sampling.

1.4 Terminology & notation

(Study) Population: A (study) population is a collection of individuals or observations about which we are interested. We mathematically denote the population’s size using upper case N. In our simulations the (study) population was the collection of N = 2400 identically sized red and white balls contained in the bowl.
Population parameter: A population parameter is a numerical summary quantity about the population that is unknown, but you wish you knew. For example, when this quantity is a mean, the population parameter of interest is the population mean which is mathematically denoted with the Greek letter \(\mu\) (pronounced “mu”). In our simulations however since we were interested in the proportion of the bowl’s balls that were red, the population parameter is the population proportion which is mathematically denoted with the letter \(p\).
Census: An exhaustive enumeration or counting of all \(N\) individuals or observations in the population in order to compute the population parameter’s value exactly. In our simulations, this would correspond to manually going over all \(N = 2400\) balls in the bowl and counting the number that are red and computing the population proportion \(p\) of the balls that are red exactly. When the number \(N\) of individuals or observations in our population is large, as was the case with our bowl, a census can be very expensive in terms of time, energy, and money.
Sampling: Sampling is the act of collecting a sample from the population when we don’t have the means to perform a census. We mathematically denote the sample’s size using lower case \(n\), as opposed to upper case \(N\) which denotes the population’s size. Typically the sample size \(n\) is much smaller than the population size \(N\), thereby making sampling a much cheaper procedure than a census. In our simulations, we used shovels with 25, 50, and 100 slots to extract a sample of size \(n = 25\), \(n = 50\), and \(n = 100\) balls.
Point estimate (AKA sample statistic): A summary statistic computed from the sample that estimates the unknown population parameter. In our simulations, recall that the unknown population parameter was the population proportion and that this is mathematically denoted with p. Our point estimate is the sample proportion: the proportion of the shovel’s balls that are red. In other words, it is our guess of the proportion of the bowl’s balls balls that are red. We mathematically denote the sample proportion using \(\hat{p}\); the “hat” on top of the p indicates that it is an estimate of the unknown population proportion \(p\).
Representative sampling: A sample is said be a representative sample if it is representative of the population. In other words, are the sample’s characteristics a good representation of the population’s characteristics? In our simulations, are the samples of \(n\) balls extracted using our shovels representative of the bowl’s \(N = 2400\) balls?
Generalizability: We say a sample is generalizable if any results based on the sample can generalize to the population. In other words, can the value of the point estimate be generalized to estimate the value of the population parameter well? In our simulations, can we generalize the values of the sample proportions red of our shovels to the population proportion red of the bowl? Using mathematical notation, is \(\hat{p}\) a “good guess” of \(p\)?
Bias: In a statistical sense, we say bias occurs if certain individuals or observations in a population have a higher chance of being included in a sample than others. We say a sampling procedure is unbiased if every observation in a population had an equal chance of being sampled. In our simulations, since each ball had the same size and hence an equal chance of being sample in our shovels, our samples were unbiased.
Random sampling: We say a sampling procedure is random if we sample randomly from the population in an unbiased fashion. In our simulations, this would correspond to sufficiently mixing the bowl before each use of the shovel.

Let’s put them all together:

If we extract a sample of \(n=50\) balls at random, in other words we mix the equally-sized balls before using the shovel, then
the contents of the shovel are an unbiased representation of the contents of the bowl’s 2400 balls, thus
any result based on the sample of balls can generalize to the bowl, thus
the sample proportion \(\hat{p}\) of the \(n=50\) balls in the shovel that are red is a “good guess” of the population proportion \(p\) of the \(N =2400\) balls that are red, thus
instead of manually going over all the balls in the bowl, we can infer about the bowl using the shovel.

Definition 1.1 The sampling distribution of a Statistic (e.g. Mean, Median, Proportion, etc) is its probability distribution.

Definition 1.2 The standard deviation of a sampling distribution is called the standard error.

Example: This is the same table as above, but notice the 2nd column name.

Number of slots in shovel	Standard Error of proportions red
25	0.0978
50	0.0669
100	0.0471

Exercise 1.2 Find and plot the sampling distribution of the proportion (\(\hat{p}\)) of heads when you flip a fair coin. (Use 5000 sets of 10 tosses)

What is the sample size?
How many experiments?
Find and plot the sampling distribution of \(\hat{p}\).
Find the standard error of the sampling distribution of \(\hat{p}\).
What happens to the standard error, when you increase the sample size?

1.5 DETOUR — Some brand name distributions

1.5.1 Normal Distribution

The normal distribution is defined by the following probability density function, where \(\mu\) is the population mean and \(\sigma\) is the standard deviation.

\[f(x) = \dfrac{1}{\sigma \sqrt{2 \pi}}e^{-(x-\mu)^2/{2\sigma^2}}\]

If a random variable \(X\) follows the normal distribution, then we write: \(X \sim N(\mu, \sigma^2)\)

Here is how the normal density looks like: Ex: here \(X \sim N(0, 1)\)

# Ignore this code

p1 <- ggplot(data = data.frame(x = c(-3, 3)), aes(x)) +
  stat_function(fun = dnorm, n = 101, args = list(mean = 0, sd = 1)) + ylab("") +
  scale_y_continuous(breaks = NULL)
p1

R functions: (Package: stats)

Examples:

Generate a random sample of 100 from \(N(15, 9)\) and create a histrogram.

x <- rnorm(n = 100, mean = 15, sd = 3)

ggplot(data.frame(x), aes(x = x)) + geom_histogram(binwidth = 1.5)

If \(X \sim N(15, 9)\) find the probability that X being greater than 21: \(P(X > 21)\)

pnorm(21, mean = 15, sd = 3) # pnorm gives us the left tail area to a given number, 21 in this case

[1] 0.9772499

If \(X \sim N(15, 9)\) find the 25th quantile.

qnorm(.25, mean = 15, sd = 3)

[1] 12.97653

#pnorm(12.97653, mean = 15, sd = 3)

Question:

A radar unit is used to measure speeds of cars on a motorway. The speeds are N(90 km/hr, 10 km/hr). What is the probability that a car picked at random is travelling at more than 100 km/hr?
GMAT are roughly normally distributed with a mean of 527 and a standard deviation of 112. How high must an individual score on the GMAT in order to score in the highest 5%?

1.5.2 Exponential Distribution

The Exponential Distribution is defined by the following probability density function, where \(\dfrac{1}{\lambda}\) is the population mean and standard deviation.

\[f(x) = \lambda e^{-\lambda x}\]

If a random variable \(X\) follows the Exponential Distribution, then we write: \(X \sim Exp(\lambda)\)

Here is how the Exponential density looks like: Ex: here \(X \sim Exp(1/15)\)

# Ignore this code
x <- seq(0, 100, length.out=1000)
dat <- data.frame(x=x, px=dexp(x, rate=1/15))

ggplot(dat, aes(x=x, y=px)) + geom_line()

R functions: (Package: stats)

Examples:

Generate a random sample of 100 from \(Exp(1/15)\) and create a histrogram.

x <- rexp(n = 100, rate = 1/15)

ggplot(data.frame(x), aes(x = x)) + geom_histogram(binwidth = 5)

If \(X \sim Exp(1/15)\) find the probability that X being less than 21: \(P(X > 21)\)

pexp(21, rate = 1/15)

[1] 0.753403

If \(X \sim Exp(1/15)\) find the 75th quantile.

qexp(.75, rate = 1/15)

[1] 20.79442

#pexp(20.79442, rate = 1/15)

Question:

The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days.

Find the probability that a traveler will purchase a ticket fewer than ten days in advance.
How many days do 80% of all travelers wait?

1.5.3 Binomial Distribution

The binomial distribution is a discrete probability distribution. It describes the outcome of n independent trials in an experiment. Each trial is assumed to have only two outcomes, either success or failure. If the probability of a successful trial is p, then the probability of having x successful outcomes in an experiment of n independent trials is as follows.

\[f(x) = {n \choose x} p^x (1-p)^{(n-x)} \quad \text{where x = 0, 1, 2,...,n}\]

Example: Suppose there are twelve multiple choice questions in an English class quiz. Each question has five possible answers, and only one of them is correct.

Find the probability of having exactly four correct answers if a student attempts to answer every question at random.
Find the probability of having four or less correct answers if a student attempts to answer every question at random.

Solution:

Since only one out of five possible answers is correct, the probability of answering a question correctly by random is 1/5=0.2.

By hand:

\({12 \choose 4} 0.2^4 (1-0.2)^{(12-4)} = 0.1329\)

In R:

dbinom(4, size=12, prob=0.2)

[1] 0.1328756

By hand:

\({12 \choose 4} 0.2^4 (1-0.2)^{(12-4)} + {12 \choose 3} 0.2^3 (1-0.2)^{(12-3)} + {12 \choose 2} 0.2^2 (1-0.2)^{(12-2)} + {12 \choose 1} 0.2^1 (1-0.2)^{(12-1)} + {12 \choose 0} 0.2^0 (1-0.2)^{(12-0)} = 0.9274\)